The proof essentially follows the outline of the proof in [1]} for Schrödinger operators on \(\mathbb {R}^n\)  (see also [2]}).
Let \(0\le V\in L_{n/2}(n_\theta )\) , and set \(A=\dot{\Delta }-\dot{\lambda }(V)\) . Let \(\lbrace u_0, \ldots , u_N\rbrace \subseteq \dot{W}_2^{1}(n_\theta )\)  be an orthonormal family in \(L_2(n_\theta )\) . Note that the \(u_\ell \)  are in the domain of the quadratic form \(Q_A\) , since \(Q_A\)  is the restriction to \(\dot{W}_2^{1}(n_\theta )\)  of \(Q_\Delta +Q_{-V}\) . Let \(\Pi = \sum _\ell \langle u_\ell |u_\ell \rangle \)  be the orthogonal projection onto \(\operatorname{Span}\lbrace u_0, \ldots , u_N\rbrace \) . We have
\(\sum _{\ell =0}^N Q_A(u_\ell ,u_\ell )\ge \sum _{\ell =0}^N -Q_{A_-}(u_\ell ,u_\ell )= \operatorname{Tr}\big [\Pi A_{-}\Pi \big ]=-\sum _{j\ge 0}\mu _j(\Pi A_{-}\Pi ).\) 
As \(\mu _j(\Pi A_{-}\Pi )\le \mu _j(A_{-})=|\lambda ^{-}_j(A)|\) , we get
\(\sum _{\ell =0}^N Q_A(u_\ell ,u_\ell )\ge - \sum _{j\ge 0} |\lambda ^{-}_j(A)|.\) 
This inequality is an instance of the variational principle for sums of eigenvalues (see, e.g., [3]}).
We also have
\(Q_A(u_\ell ,u_\ell ) & = \left\langle \Delta u_\ell  , u_\ell  \right\rangle  - \left\langle \lambda (V)u_\ell  , u_\ell  \right\rangle \\& = -\sum _{j=1}^n \left\langle \partial _j^2u_\ell  , u_\ell  \right\rangle  -\tau \big [ V|u_\ell |^2\big ]\\&= \tau \big [|\nabla u_\ell |^2\big ] -\tau \big [ V|u_\ell |^2\big ].\) 

Here \(|\nabla u_\ell |^2=\sum |\partial _j u_\ell |^2\in L_1(n_\theta )\) . Moreover, as \(W_2^1(n_\theta )\subseteq L_{\frac{2n}{n-2}}(n_\theta )\)  by Proposition REF , we see that \(|u_\ell |^2\in L_{\frac{n}{n-2}}(n_\theta )\) , and hence \(V|u_\ell |^2\in L_1(n_\theta )\) . Thus, if we set \(\rho =\sum |u_\ell |^2\in L_{\frac{n}{n-2}}(n_\theta )\) , then we obtain
\(\sum _{\ell =0}^N Q_A(u_\ell ,u_\ell ) = \sum _{\ell =0}^N \tau \big [|\nabla u_\ell |^2\big ] - \tau \big [ \rho V\big ].\) 
If we further assume that \(V\in L_{\frac{n}{2}+1}(n_\theta )\) , then the LT inequality (REF ) for \(p=n/2\)  and \(\gamma =1\)  delivers
\(\sum _{j\ge 0} |\lambda ^{-}_j(A)| \le L_n \tau \big [V^{\frac{n}{2}+1}\big ].\) 
Combining this with (REF ) and (REF ) shows that, for all \(0\le V\in L_{1+n/2}(n_\theta )\) , we have
\(\sum _{\ell =0}^N \tau \big [|\nabla u_\ell |^2\big ] \ge \tau \big [\rho V - L_n V^{\frac{n}{2}+1}\big ].\) 
The original argument of Lieb-Thirring relies on the following observation. Given \(a,b,\alpha >0\) , the function \(\varphi (t)=at-bt^{\alpha +1}\) , \(t>0\) , reaches its maximum at \(t_0=[(\alpha +1)b]^{-1/\alpha }a^{1/\alpha }\)  and the maximum value then is equal to
\(\varphi (t_0)= t_0(a-bt_0^\alpha )= at_0\big (1 -\frac{1}{\alpha +1}\big )= \frac{\alpha }{\alpha +1}\big [(\alpha +1)b\big ]^{-\frac{1}{\alpha }}a^{1+\frac{1}{\alpha }}>0 .\) 
Likewise, for \(\alpha =n/2\) , \(a=\rho \)  and \(b=L_n\) , by taking \(V=(\frac{1}{2}(n+2)L_n)^{-2/n}\rho ^{2/n}\in L_{\frac{n^2}{2(n-2)}}(n_\theta )\subseteq L_{\frac{n}{2}+1}(n_\theta )\) , we have \(\rho V - L_n V^{\frac{n}{2}+1}=\frac{n}{n+2}(\frac{1}{2} (n+2)L_n)^{-2/n}\rho ^{(n+2)/n}\) . In this case (REF ) gives
\(\sum _{\ell =0}^N \tau \big [|\nabla u_\ell |^2\big ] \ge \frac{n}{n+2}\bigg (\frac{n+2}{2}L_n\bigg )^{-\frac{2}{n}}\tau \big [\rho ^{\frac{n+2}{n}}\big ].\) 
This shows that the Sobolev inequality (REF ) holds with a best constant \(K_n\)  such that
\(K_n\ge \frac{n}{n+2}\bigg (\frac{n+2}{2}L_n\bigg )^{-\frac{2}{n}}.\) 
Conversely, let \(V=V^*\in L_{\frac{n}{2}+1}(n_\theta )\) , and set \(\dot{H}_{V}=\dot{\Delta }+\dot{\lambda }(V)\) . Set \(N=N^{-}(H_V)-1\) , and let \(\lbrace u_0, \ldots , u_N\rbrace \subseteq \dot{W}_2^{1}(n_\theta )\)  be an orthonormal family such that \(\dot{H}_Vu_\ell = \lambda ^{-}_{\ell }(H_V)u_\ell \)  for \(\ell =0, \ldots , N\) . We have
\(\sum _{\ell =0}^N \left|\lambda ^{-}\big (\dot{H}_V\big )\right|= -\sum _{\ell =0}^N Q_{\dot{H}_V}(u_\ell ,u_\ell ) \le -\sum _{\ell =0}^N Q_{\dot{H}_{-V_{-}}}(u_\ell ,u_\ell ).\) 
As above, set \(\rho = \sum |u_\ell |^2\in L_{\frac{n}{n-2}}(n_\theta )\) . By using (REF ) and the Sobolev inequality (REF ) we get
\(\sum _{\ell =0}^N \left|\lambda ^{-}\big (\dot{H}_V\big )\right| \le -\sum _{\ell =0}^N \tau \big [|\nabla u_\ell |^2\big ] + \tau \big [ \rho V_{-}\big ] \le -K_n \tau \big [\rho ^{\frac{n+2}{n}}\big ] + \tau \big [ \rho V\big ].\) 
Set \(q=1+n/2\)  and \(r=1+2/n\) , so that \(q^{-1}+r^{-1}=1\) . Recall that \(V\in L_{q}(n_\theta )\)  and \(\rho \in L_{\frac{n}{n-2}}(n_\theta )\subseteq L_r(n_\theta )\) . Thus, by Hölder's inequality,
\(\tau \big [ \rho V_{-}\big ] \le \tau \big [|V_{-}|^q\big ]^{\frac{1}{q}} \tau \big [\rho ^r\big ]^{\frac{1}{r}}.\) 
Recall the inequality \(ab\le q^{-1}a^q + r^{-1}b^r\)  for \(a,b>0\)  (this is a convexity inequality for the function \(t\rightarrow a^{1-t}b^t\) ). Applying it to \(a=\epsilon \tau \big [|V_{-}|^q\big ]^{\frac{1}{q}} \)  and \(b=\epsilon ^{-1}\tau \big [\rho ^r\big ]^{\frac{1}{r}}\)  with \(\epsilon >0\)  gives
\(\tau \big [ \rho V_{-}\big ] \le \frac{1}{p}\epsilon ^q\tau \big [|V_{-}|^q\big ] + \frac{1}{r} \epsilon ^{-r} \tau \big [\rho ^r\big ].\) 
Combining this with (REF ) gives
\(\sum _{\ell =0}^N \left|\lambda ^{-}\big (\dot{H}_V\big )\right|\le \bigg (\frac{1}{r} \epsilon ^{-r}-K_n\bigg ) \tau \big [\rho ^{r}\big ] + \frac{1}{q}\epsilon ^q\tau \big [|V_{-}|^q\big ].\) 
If we choose \(\epsilon \)  so that \(r^{-1} \epsilon ^{-r}=K_n\) , i.e., \(\epsilon = (rK_n)^{-1/r}\) , then we get
\(\sum _{\ell =0}^N \left|\lambda ^{-}\big (\dot{H}_V\big )\right|\le \frac{1}{q}\epsilon ^q\tau \big [|V_{-}|^q\big ] \qquad \text{for all}\ V=V^*\in L_{\frac{n}{2}+1}(n_\theta ).\) 
This reproves the LT inequality (REF ) for \(p=n/2\)  and \(\gamma =1\)  and shows that the best LT constant \(L_n\)  satisfies
\(L_n \le \frac{1}{q}\epsilon ^q = \frac{1}{q} (rK_n)^{-\frac{q}{r}}=\frac{2}{n+2}\bigg ( \frac{n}{n+2}K_n\bigg )^{-\frac{n}{2}}.\) 
Combining this with (REF ) gives (REF ). The proof is complete.
